Physics Circle home Student Login

Home›Chapters›Moving Charges and Magnetism PYQs›Force on Current-Carrying Conductors›Question 41

JEE Main 2022 (28 Jul, Shift 1) · Force on Current-Carrying Conductors

Force on Current-Carrying Conductors question from JEE Main 2022 (28 Jul, Shift 1)

Q41JEE Main 2022 · 28 JUL · SHIFT 1Medium

Force on Current-Carrying Conductors

As shown in the figure, a metallic rod of linear density 0.45 kg m−10.45\text{ kg m}^{-1} is lying horizontally on a smooth inclined plane which makes an angle of 45∘45^\circ with the horizontal. The minimum current flowing in the rod required to keep it stationary, when 0.15 T0.15\text{ T} magnetic field is acting on it in the vertical upward direction, will be:

{Use g=10 m/s2g = 10\text{ m/s}^2}

Figure

Show answer and solution

Answer: (a)

  1. Vertical BB gives a horizontal force F=IlBF = I l B on the rod.
  2. Along the smooth incline: IlBcos⁡θ=mgsin⁡θ⇒I=(m/l) gtan⁡θBI l B\cos\theta = mg\sin\theta \Rightarrow I = \frac{(m/l)\,g\tan\theta}{B}
  3. I=0.45×10×tan⁡45∘0.15=30 AI = \frac{0.45\times10\times\tan45^\circ}{0.15} = 30\text{ A}
  4. So the answer is (a) 30 A.
Report a mistake

More questions on Force on Current-Carrying Conductors